16.1C: Role of Microbes in Biogeochemical Cycling
- Explain the role microbes play in biogeochemical cycling
Microbial Role in Biogeochemical Cycling
Nutrients move through the ecosystem in biogeochemical cycles. A biogeochemical cycle is a pathway by which a chemical element (such as carbon or nitrogen) circulates through the biotic (living) and the abiotic (non-living) factors of an ecosystem. The elements that move through the factors of an ecosystem are not lost but are instead recycled or accumulated in places called reservoirs (or “sinks”) where they can be held for a long period of time. Elements, chemical compounds, and other forms of matter are passed from one organism to another and from one part of the biosphere to another through these biogeochemical cycles.
Ecosystems have many biogeochemical cycles operating as a part of the system. A good example of a molecule that is cycled within an ecosystem is water, which is always recycled through the water cycle. Water undergoes evaporation, condensation, and then falls back to Earth as rain (or other forms of precipitation). This typifies the cycling that is observed for all of the principal elements of life.
Although biogeochemical cycles in a given ecosystem are coordinated by the full complement of living organisms and abiotic factors that make up that system, microorganisms play a primary role in regulating biogeochemical systems in virtually all of our planet’s environments. This includes extreme environments such as acid lakes and hydrothermal vents, and even includes living systems such as the human gut. The key collective metabolic processes of microbes (including nitrogen fixation, carbon fixation, methane metabolism, and sulfur metabolism) effectively control global biogeochemical cycling. Incredibly, production by microbes is so immense that global biogeochemistry would likely not change even if eukaryotic life were totally absent!
Microbes comprise the backbone of every ecological system, particularly those in which there is no light (i.e. systems in which energy cannot be collected through photosynthesis ). Two key examples of critical biogeochemical processes carried out by microorganisms are discussed below.
The Carbon Cycle
Carbon is critical for life because it is the essential building block of all organic compounds. Plants and animals utilize carbon to produce carbohydrates, fats, and proteins, which can then be used to build their internal structures or to obtain energy.
Carbon in the form of carbon dioxide (CO 2 ) is readily obtained from the atmosphere, but before it can be incorporated into living organisms it must be transformed into a usable organic form. The transformative process by which carbon dioxide is taken up from the atmospheric reservoir and “fixed” into organic substances is called carbon fixation. Perhaps the best known example of carbon fixation is photosynthesis, a process by which energy derived from sunlight is harnessed to form organic compounds. Photosynthesis depends on the activity of microorganisms such as cyanobacteria; indeed, the fact that there is oxygen in the Earth’s atmosphere at all is a consequence of the photosynthetic activity of ancient microbes.
The Nitrogen Cycle
Nitrogen is essential for all forms of life because it is required for synthesis of the basic building blocks of life (e.g., DNA, RNA, and amino acids). The Earth’s atmosphere is primarily composed of nitrogen, but atmospheric nitrogen (N 2 ) is relatively unusable for biological organisms. Consequently, chemical processing of nitrogen (or nitrogen fixation) is necessary to convert gaseous nitrogen into forms that living organisms can use. Almost all of the nitrogen fixation that occurs on the planet is carried out by bacteria that have the enzyme nitrogenase, which combines N 2 with hydrogen to produce a useful form of nitrogen (such as ammonia). Thus, microorganisms are absolutely essential for plant and animal life forms, which cannot fix nitrogen on their own.
Key Points
- A biogeochemical cycle is a pathway by which a chemical element (such as carbon or nitrogen) circulates through and is recycled by an ecosystem.
- Microorganisms play a primary role in regulating biogeochemical systems in virtually all of our planet ‘s environments.
- Microbes participate in essential biogeochemical cycling events such as carbon and nitrogen fixation.
Key Terms
- photosynthesis : The process by which plants and other photoautotrophs generate carbohydrates and oxygen from carbon dioxide, water, and light energy in chloroplasts.
- biogeochemistry : The scientific study of biological, geological, and chemical processes in the natural environment and especially of their mutual relationships.
- nitrogenase : The enzyme, in nitrogen-fixing bacteria, that catalyzes the conversion of atmospheric nitrogen into ammonia.


6.1: Chapter 10
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Chapter 10 Critical Thinking Activities
You can review additional information regarding these answers in the corresponding section in which the Critical Thinking activities appear.
Critical Thinking Activity Section 10.6a
The patient should be advised that acetaminophen can cause acute liver damage when taken in excessive amounts or when used with alcohol. Many over-the-counter medications contain acetaminophen, so daily amounts must be monitored carefully. Recommended daily restrictions for acetaminophen include less than 4,000 mg of acetaminophen in 24 hours for an adult, less than 3200 mg for geriatric adults, and less than 2000 mg for patients with alcoholism. Fewer than three alcoholic drinks should be consumed daily while using acetaminophen.
Critical Thinking Activity Section 10.6b
The patient should be advised that aspirin has an anti-platelet effect, in addition to reducing pain, fever, and inflammation. By preventing the platelets from sticking together, clots that can cause heart attacks and strokes are prevented from forming.
Critical Thinking Activity Section 10.6c
Ibuprofen is a nonsteroidal anti-inflammatory drug (NSAID), which can cause severe and life-threatening stomach bleeding and must be taken cautiously. The patient should be advised that the risk for bleeding is higher if the patient:
- is age 60 or older
- has had stomach ulcers or bleeding problems
- takes a anticoagulant or steroid medication
- takes other drugs containing NSAIDs (such as aspirin, ibuprofen, or naproxen)
- consumes three or more alcoholic drinks every day while using this product
- takes ibuprofen in higher doses, more frequently, or for a longer time than directed
Critical Thinking Activity Section 10d
The nurse should evaluate the effectiveness of ketorolac IV in relieving the patient’s pain 30 minutes after administration.
Critical Thinking Activity Section 10e
The nurse should provide the following patient education to a patient who has been prescribed celecoxib:
- It may be taken with or without food
- You can sprinkle capsules on applesauce and ingest it immediately with water
- You may experience heartburn, vomiting, or diarrhea with this medication
- Notify the provider immediately if you have abdominal pain, vomit blood or have blood in your stool, develop swelling in your hand or feet, or notice yellowing of your skin
Critical Thinking Activity Section 10.7a
Oral drops of morphine, commonly used for patients with metastatic cancer, should be effective within 1 hour of administration.
Critical Thinking Activity Section 10.7b
Naloxone immediately reverses the effects of respiratory depression and oversedations caused by opioids. After a patient receives naloxone, the nurse should continue to evaluate the patient’s respiratory status at least every 15 minutes because naloxone has a shorter duration of action that many opioids, and repeated doses are usually necessary.
Critical Thinking Activity Section 10.8a
The nurse should educate the patient to take baclofen with milk or food to minimize gastric upset. Advise the patient that baclofen may cause dizziness or drowsiness, so they should change positions slowly and avoid driving and operating machines. Patients using baclofen should avoid using alcohol or taking other CNS depressants.
Critical Thinking Activity Section 10.8b
Cyclobenzaprine is a muscle relaxer and may cause drowsiness. If used with alcohol or other CNS depressants, it can impair mental or physical abilities, so the patient should be advised to not drive when taking cyclobenzaprine.
Critical Thinking Activity Section 10.8c
The patient should be advised that tizanidine can cause drowsiness and dizziness, and concurrent use with alcohol can worsen these effects. Tizanidine can damage the liver so alcohol should be avoided to prevent additional damage.
Critical Thinking Activity Section 10.11
The correct answers are a), b), and e). Based on the patient’s respiratory status, the nurse should immediately raise the patient’s bed and apply oxygen to rapidly increase their oxygenation level. The nurse should ask for help from a team member and/or call the rapid response team while obtaining naloxone to administer for sedation and respiratory depression. The nurse should continue to monitor the patient’s respiratory status after naloxone is administered because repeated doses may be required.


7.3C: Regional Vertebral Characteristics
The vertebrae of the spinal column are divided into five regions: cervical, thoracic, lumbar, sacral, and coccyx.
- Identify the different types of vertebrae
Key Points
- The upper three regions of the vertebral column are articulating vertebrae grouped under the names cervical, thoracic and lumbar, according to the regions they occupy.
- Vertebrae are given an alphanumeric descriptor, with the initial letter derived from the region they are located in followed by a digit that increases moving down the region. For example, the most superior cervical vertebrae is termed C1 and the most inferior C7, which is then followed by the T1 vertebrae of the thoracic region.
- The two lower most regions of vertebrae are the sacrum and the coccyx.
The vertebrae comprising the spinal column can be divided into five regions, based on the five varying curvatures of the spine. The upper three regions of the spinal column are termed the cervical, thoracic, and lumbar; they contain individually jointed vertebrae. The two lower regions—the sacrum and coccyx, or tailbone—are formed from fused vertebrae.
Upper Vertebrae
Vertebrae are given an alphanumeric descriptor, with the initial letter derived from the region they are located in followed by a digit that increases moving down the region. For example, the most superior cervical vertebrae is termed C1 and the most inferior C7, which is then followed by the T1 vertebrae of the thoracic region.
The cervical region of the spine is the most superior and contains seven small vertebrae. The main function of the cervical region is to facilitate attachment of the skull to the spine, protect the spinal cord over the exposed neck and shoulder region, and support the body.
Cervical vertebra, lateral view : The lateral view of a typical cervical vertebra.
The twelve thoracic vertebrae are located inferiorly to the cervical region. They are larger than the cervical vertebrae and increase in size moving inferiorly to the lumbar region.
Thoracic vertebra : Image of a typical thoracic vertebra.
The five lumbar vertebrae are the largest vertebral bones and increase in size when moving inferiorly. The lumbar vertebrae play a key role in supporting the body and facilitating locomotion.
Lumbar vertebra : Image of typical lumbar vertebra.
Lower Vertebrae
During childhood the five vertebrae of the sacral region are distinct. In adulthood the five bones fuse to form the sacrum, although it is still often divided into regions termed S1–S5 based on the formation of the original individual bones. The sacrum functions to support the body and protect organs of the pelvis and lower back.
The final region of the spine is the coccyx, or tailbone. As with the sacrum, the coccyx is formed from several vertebrae that have fused together.
As it’s alternative name suggests, the coccyx forms the basis of a tail that has been lost in humans, although it is incorrect to think of it as a vestigial structure since is a key attachment point for many muscles and ligaments and plays a key role in supporting the body while sitting.
Coccyx : Lateral view of coccyx shown beneath the sacrum.
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7.1: Introduction to the Appendicular Skeleton
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Because of our upright stance, different functional demands are placed upon the upper and lower limbs. Thus, the bones of the lower limbs are adapted for weight-bearing support and stability, as well as for body locomotion via walking or running. In contrast, our upper limbs are not required for these functions. Instead, our upper limbs are highly mobile and can be utilized for a wide variety of activities.
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7.2: The Pectoral Girdle
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The appendicular skeleton includes all of the limb bones, plus the bones that unite each limb with the axial skeleton. The bones that attach each upper limb to the axial skeleton form the pectoral girdle (shoulder girdle). This consists of two bones, the scapula and clavicle. The clavicle (collarbone) is an S-shaped bone located on the anterior side of the shoulder. It is attached on its medial end to the sternum of the thoracic cage, which is part of the axial skeleton.
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7.3: Bones of the Upper Limb
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The upper limb is divided into three regions. These consist of the arm, located between the shoulder and elbow joints; the forearm, which is between the elbow and wrist joints; and the hand, which is located distal to the wrist. There are 30 bones in each upper limb. The humerus is the single bone of the upper arm, and the ulna (medially) and the radius (laterally) are the paired bones of the forearm.
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7.4: The Pelvic Girdle and Pelvis
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The pelvic girdle (hip girdle) is formed by a single bone, the hip bone or coxal bone (coxal = “hip”), which serves as the attachment point for each lower limb. Each hip bone, in turn, is firmly joined to the axial skeleton via its attachment to the sacrum of the vertebral column. The right and left hip bones also converge anteriorly to attach to each other. The bony pelvis is the entire structure formed by the two hip bones, the sacrum, and, attached inferiorly to the sacrum, the coccyx .
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7.5: Bones of the Lower Limb
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Like the upper limb, the lower limb is divided into three regions. The thigh is that portion of the lower limb located between the hip joint and knee joint. The leg is specifically the region between the knee joint and the ankle joint. Distal to the ankle is the foot. The lower limb contains 30 bones. These are the femur, patella, tibia, fibula, tarsal bones, metatarsal bones, and phalanges.
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7.6: Embryonic Development of the Appendicular Skeleton
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Embryologically, the appendicular skeleton arises from mesenchyme, a type of embryonic tissue that can differentiate into many types of tissues, including bone or muscle tissue. Mesenchyme gives rise to the bones of the upper and lower limbs, as well as to the pectoral and pelvic girdles. Development of the limbs begins near the end of the fourth embryonic week, with the upper limbs appearing first. Thereafter, the development of the upper and lower limbs follows similar patterns.


3: Real Number Systems and Number Theory Last updated Save as PDF Page ID 129479 OpenStax OpenStax 3.0: Introduction 3.1: Prime and Composite Numbers 3.1.0: Exercises 3.2: The Integers 3.2.0: Exercises 3.3: Order of Operations 3.3.0: Exercises 3.4: Rational Numbers 3.4.0: Exercises 3.5: Irrational Numbers 3.5.0: Exercises 3.6: Real Numbers 3.6.0: Exercises 3.7: Clock Arithmetic 3.7.0: Exercises 3.8: Exponents 3.8.0: Exercises 3.9: Scientific Notation 3.9.0: Exercises 3.10: Arithmetic Sequences 3.10.0: Exercises 3.11: Geometric Sequences 3.11.0: Exercises 3.12: Chapter Summary 3.12.0: Key Terms 3.12.1: Key Concepts 3.12.2: Videos 3.12.3: Formula Review 3.12.4: Projects 3.12.5: Chapter Review 3.12.6: Chapter Test


4.6: Atomic Masses
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Learning Objectives
- Define atomic mass and atomic mass unit.
- Calculate atomic mass.
Even though atoms are very tiny pieces of matter, they have mass. Their masses are so small, however, that chemists often use a unit other than grams to express them—the atomic mass unit.
Atomic Mass Unit
The atomic mass unit (abbreviated u, altho ugh amu is a lso used) is defined as 1/12 of the mass of a 12 C atom:
\[\mathrm{1\:u=\dfrac{1}{12}\textrm{ the mass of }^{12}C\:atom} \label{Eq1} \]
It is equal to 1.661 × 10 −24 g.
Masses of other atoms are expressed with respect to the atomic mass unit. For example, the mass of an atom of 1 H is 1.008 u, the mass of an atom of 16 O is 15.995 u, and the mass of an atom of 32 S is 31.97 u. Note, however, that these masses are for particular isotopes of each element. Because most elements exist in nature as a mixture of isotopes, any sample of an element will actually be a mixture of atoms having slightly different masses (because neutrons have a significant effect on an atom’s mass). How, then, do we describe the mass of a given element? By calculating an average of an element’s atomic masses, weighted by the natural abundance of each isotope, we obtain a weighted average mass called the atomic mass (also commonly referred to as the atomic weight ) of an element.
Atomic Mass is the Weighted Average Mass of Isotopes
As stated above, most elements occur naturally as a mixture of two or more isotopes. Listed below (Table \(\PageIndex{1}\)) are the naturally occurring isotopes of selected elements along with the percent natural abundance of each.
| Element | Isotope (Symbol) | Percent Natural Abundance | Atomic Mass \(\left( \text{amu} \right)\) | Average Atomic Mass \(\left( \text{amu} \right)\) |
|---|---|---|---|---|
| Hydrogen | \(\ce{_1^1H}\) | 99.985 | 1.0078 | 1.0079 |
| \(\ce{_1^2H}\) | 0.015 | 2.0141 | ||
| \(\ce{_1^3H}\) | negligible | 3.0160 | ||
| Carbon | \(\ce{_6^{12}C}\) | 98.89 | 12.000 | 12.011 |
| \(\ce{_6^{13}C}\) | 1.11 | 13.003 | ||
| \(\ce{_6^{14}C}\) | trace | 14.003 | ||
| Oxygen | \(\ce{_8^{16}O}\) | 99.759 | 15.995 | 15.999 |
| \(\ce{_8^{17}O}\) | 0.037 | 16.995 | ||
| \(\ce{_8^{18}O}\) | 0.204 | 17.999 | ||
| Chlorine | \(\ce{_{17}^{35}Cl}\) | 75.77 | 34.969 | 35.453 |
| \(\ce{_{17}^{38}Cl}\) | 24.23 | 36.966 | ||
| Copper | \(\ce{_{29}^{63}Cu}\) | 69.17 | 62.930 | 63.546 |
| \(\ce{_{29}^{65}Cu}\) | 30.83 | 64.928 |
For some elements, one particular isotope is much more abundant than any other isotopes. For example, naturally occurring hydrogen is nearly all hydrogen-1, and naturally occurring oxygen is nearly all oxygen-16. For many other elements, however, more than one isotope may exist in substantial quantities. Chlorine (atomic number 17) is yellowish-green toxic gas. About three quarters of all chlorine atoms have 18 neutrons, giving those atoms a mass number of 35. About one quarter of all chlorine atoms have 20 neutrons, giving those atoms a mass number of 37. Were you to simply calculate the arithmetic average of the precise atomic masses, you would get approximately 36.
\[\frac{34.969 \,u + 36.966 \,u}{2} = 35.968 \,u \nonumber \]
As you can see, the average atomic mass given in the last column of the table above (35.453) is significantly lower. Why? The reason is that we need to take into account the natural abundance percentages of each isotope in order to calculate what is called the weighted average . The atomic mass of an element is the weighted average of the atomic masses of the naturally occurring isotopes of that element.
\[0.7577 \left( 34.969 \,u \right) + 0.2423 \left( 36.966 \,u \right) = 35.453 \,u \nonumber \]
The weighted average is determined by multiplying the percent of natural abundance by the actual mass of the isotope. This is repeated until there is a term for each isotope. For chlorine, there are only two naturally occurring isotopes so there are only two terms.
Atomic mass = (%1)(mass 1) + (%2)(mass 2) + ⋯
Another example: oxygen exists as a mixture that is 99.759% 16 O, 0.037% 17 O and 0.204% 18 O. The atomic mass of oxygen (use percent natural abundance data from Table 2.5.1) would be calculated as follows:
Atomic mass = (%1)(mass 1) + (%2)(mass 2) + (%3)(mass 3)
\[0.99759 \left( 15.995 u \right) + 0.00037 \left( 16.995 u \right) +0.00204 \left( 17.999 u \right)= 15.999 u \nonumber \]
To confirm your answer, compare the calculated value to the weighted mass displayed on the periodic table.
Example \(\PageIndex{1}\)
Calculate the atomic mass of oxygen. Oxygen exists as a mixture of 3 isotopes. Their respective masses and natural abundance are shown below.
- 16 O: 15.995 u (99.759%)
- 17 O: 16.995 u (0.037%)
- 18 O: 17.999 u (0.204%)
Solution
Multiply the isotope abundance by the actual mass of the isotope, and then sum up the products.
\[0.99759 \left( 15.995\, u \right) + 0.00037 \left( 16.995 \,u \right) +0.00204 \left( 17.999\, u \right)= 15.999\, u \nonumber \]
Exercise \(\PageIndex{1}\)
Calculate the atomic mass of copper. Copper exists as a mixture of 2 isotopes. Their respective masses and natural abundance are shown below.
- 63 Cu: 62.930 u (69.17%)
- 65 Cu: 64.928 u (30.83%)
- Answer
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63.546 u
The atomic mass of each element is found under the element symbol in the periodic table. Examples are shown below. The atomic mass of tin (Sn) is 118.71 u while the atomic mass of carbon (C) is 12.011 u. On the other hand, the atomic number (Z) of each element is found above the atomic symbol.
Atomic mass indicated on entries of the Periodic Table. (public Domain; Pubchem )
The periodic table is found in this link:
Example \(\PageIndex{2}\): Mass of Carbon
What is the average mass of a carbon atom in grams? The atomic mass is found in the Periodic Table. Please use two decimal places.
Solution
This is a simple one-step conversion, similar to conversions we did in Chapter 1 . We use the fact that 1 u = 1.661 × 10 −24 g:
\(\mathrm{12.01\:\cancel{u}\times\dfrac{1.661 \times 10^{-24}\:g}{1\:\cancel{u}}=1.995\times 10^{-23}\:g}\)
This is an extremely small mass, which illustrates just how small individual atoms are.
Exercise \(\PageIndex{2}\): Mass of Tin
What is the average mass of a tin atom in grams? The average atomic mass is found in the Periodic Table. Please use two decimal places.
- Answer
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\(\mathrm{118.71\:\cancel{u}\times\dfrac{1.661\times10^{-24}\:g}{1\:\cancel{u}}=1.972 \times 10^{-22}\:g}\)
Concept Review Exercises
- Define atomic mass. Why is it considered a weighted average?
- What is an atomic mass unit?
Answers
- The atomic mass is an average of an element’s atomic masses, weighted by the natural abundance of each isotope of that element. It is a weighted average because different isotopes have different masses.
- An atomic mass unit is 1/12th of the mass of a 12 C atom.
Key Takeaway
- Atoms have a mass that is based largely on the number of protons and neutrons in their nucleus.
- The atomic mass of each element in the Periodic Table is the weighted average of the mass of all its isotopes.


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7.5.E: Problems on Premeasures and Related Topics
Fill in the missing details in the proofs, notes, and examples of this section.
Describe \(m^{*}\) on \(2^{S}\) induced by a premeasure \(\mu : \mathcal{C} \rightarrow E^{*}\) such that each of the following hold.
(a) \(\mathcal{C}=\{S, \emptyset\}, \mu S=1\).
(b) \(\mathcal{C}=\{S, \emptyset, \text {and all singletons}\}; \mu S=\infty, \mu\{x\}=1\).
(c) \(\mathcal{C}\) as in (b), with \(S\) uncountable; \(\mu S=1,\) and \(\mu X=0\) otherwise.
(d) \(\mathcal{C}=\{\text {all proper subsets of } S\}; \mu X=1\) when \(\emptyset \subset X \subset S; \mu \emptyset=0\).
Show that the premeasures
\[v^{\prime} : \mathcal{C}^{\prime} \rightarrow[0, \infty]\]
induce one and the same (Lebesgue) outer measure \(m^{*}\) in \(E^{n},\) with \(v^{\prime}=v\) (volume, as in §2):
(a) \(\mathcal{C}^{\prime}=\{\text {open intervals}\}\);
(b) \(\mathcal{C}^{\prime}=\{\text {half-open intervals}\}\);
(c) \(\mathcal{C}^{\prime}=\{\text {closed intervals}\}\);
(d) \(\mathcal{C}^{\prime}=\mathcal{C}_{\sigma}\);
(e) \(\mathcal{C}^{\prime}=\{\text {open sets}\}\);
(f) \(\mathcal{C}^{\prime}=\{\text {half-open cubes}\}\);
[Hints: (a) Let \(m^{\prime}\) be the \(v^{\prime}\)-induced outer measure; let \(\mathcal{C}=\{\text {all intervals}\}.\) As \(\mathcal{C}^{\prime} \subseteq \mathcal{C}, m^{\prime} A \geq m^{*} A.\) (Why?) Also,
\[(\forall \varepsilon>0)\left(\exists\left\{B_{k}\right\} \subseteq \mathcal{C}\right) \quad A \subseteq \bigcup_{k} B_{k} \text { and } \sum v B_{k} \leq m^{*} A+\varepsilon.\]
(Why?) By Lemma 1 in §2,
\[\left(\exists\left\{C_{k}\right\} \subseteq \mathcal{C}^{\prime}\right) \quad B_{k} \subseteq C_{k} \text { and } v B_{k}+\frac{\varepsilon}{2^{k}}>v^{\prime} C_{k}.\]
Deduce that \(m^{*} A \geq m^{\prime} A, m^{*}=m^{\prime}\). Similarly for (b) and (c). For (d), use Corollary 1 and Note 3 in §1. For (e), use Lemma 2 in §2. For (f), use Problem 2 in §2.]
Do Problem 3(a)-(c), with \(m^{*}\) replaced by the Jordan outer content \(c^{*}\) (Note 6).
Do Problem 3, with \(v\) and \(m^{*}\) replaced by the LS premeasure and outer measure. (Use Problem 7 in §4.)
Show that a set \(A \subseteq E^{n}\) is bounded iff its outer Jordan content is finite.
Find a set \(A \subseteq E^{1}\) such that
(i) its Lebesgue outer measure is \(0\) \(\left(m^{*} A=0\right),\) while its Jordan outer content \(c^{*} A=\infty\);
(ii) \(m^{*} A=0, c^{*} A=1\) (see Corollary 6 in §2).
Let
\[\mu_{1}, \mu_{2} : \mathcal{C} \rightarrow[0, \infty]\]
be two premeasures in \(S\) and let \(m_{1}^{*}\) and \(m_{2}^{*}\) be the outer measures induced by them.
Prove that if \(m_{1}^{*}=m_{2}^{*}\) on \(\mathcal{C},\) then \(m_{1}^{*}=m_{2}^{*}\) on all of \(2^{S}\).
With the notation of Definition 3 and Note 6, prove the following.
(i) If \(A \subseteq B \subseteq S\) and \(m^{*} B=0,\) then \(m^{*} A=0;\) similarly for \(c^{*}\).
[Hint: Use monotonicity.]
(ii) The set family
\[\left\{X \subseteq S | c^{*} A=0\right\}\]
is a hereditary set ring, i.e., a ring \(\mathcal{R}\) such that
\[(\forall B \in \mathcal{R})(\forall A \subseteq B) \quad A \in \mathcal{R}.\]
(iii) The set family
\[\left\{X \subseteq S | m^{*} X=0\right\}\]
is a hereditary \(\sigma\)-ring.
(iv) So also is
\[\mathcal{H}=\{\text {those } X \subseteq S \text { that have basic coverings}\};\]
thus \(\mathcal{H}\) is the hereditary \(\sigma\)-ring generated by \(\mathcal{C}\) (see Problem 14 in §3).
Continuing Problem 8(iv), prove that if \(\mu\) is \(\sigma\)-finite (Definition 4), so is \(m^{*}\) when restricted to \(\mathcal{H}.\)
Show, moreover, that if \(\mathcal{C}\) is a semiring, then each \(X \in \mathcal{H}\) has a basic covering \(\left\{Y_{n}\right\},\) with \(m^{*} Y_{n}<\infty\) and with all \(Y_{n}\) disjoint.
[Hint: Show that
\[X \subseteq \bigcup_{n=1}^{\infty} \bigcup_{k=1}^{\infty} B_{n k}\]
for some sets \(B_{n k} \in \mathcal{C},\) with \(\mu B_{n k}<\infty.\) Then use Note 4 in §5 and Corollary 1 of §1.]
Show that if
\[s : \mathcal{C} \rightarrow E^{*}\]
is \(\sigma\)-finite and additive on \(\mathcal{C},\) a semiring, then the \(\sigma\)-ring \(\mathcal{R}\) generated by \(\mathcal{C}\) equals the \(\sigma\)-ring \(\mathcal{R}^{\prime}\) generated by
\[\mathcal{C}^{\prime}=\{X \in \mathcal{C}| | s X |<\infty\}\]
(cf. Problem 6 in §4).
[Hint: By \(\sigma\)-finiteness,
\[(\forall X \in \mathcal{C})\left(\exists\left\{A_{n}\right\} \subseteq \mathcal{C}| | s A_{n} |<\infty\right) \quad X \subseteq \bigcup_{n} A_{n};\]
so
\[X=\bigcup_{n}\left(X \cap A_{n}\right), \quad X \cap A_{n} \in \mathcal{C}^{\prime}.\]
(Use Lemma 3 in §4.)
Thus \((\forall X \in \mathcal{C}) X\) is a countable union of \(\mathcal{C}^{\prime}\)-sets; so \(\mathcal{C} \subseteq \mathcal{R}^{\prime}.\) Deduce \(\mathcal{R} \subseteq \mathcal{R}^{\prime}\). Proceed.]
With all as in Theorem 3, prove that if \(A\) has basic coverings, then
\[\left(\exists B \in \mathcal{A}_{\delta}\right) \quad A \subseteq B \text { and } m^{*} A=m^{*} B.\]
[Hint: By formula (4),
\[(\forall n \in N)\left(\exists X_{n} \in \mathcal{A} | A \subseteq X_{n}\right) \quad m^{*} A \leq m X_{n} \leq m^{*} A+\frac{1}{n}.\]
(Explain!) Set
\[B=\bigcap_{n=1}^{\infty} X_{n} \in \mathcal{A}_{\delta}.\]
Proceed. For \(\mathcal{A}_{\delta},\) see Definition 2(b) in §3.]
Let \((S, \mathcal{C}, \mu)\) and \(m^{*}\) be as in Definition 3. Show that if \(\mathcal{C}\) is a \(\sigma\)-field in \(S,\) then
\[(\forall A \subseteq S)(\exists B \in \mathcal{C}) \quad A \subseteq B \text { and } m^{*} A=\mu B.\]
[Hint: Use Problem 11 and Note 3.]
\(\Rightarrow^{*}\) Show that if
\[s : \mathcal{C} \rightarrow E\]
is \(\sigma\)-finite and \(\sigma\)-additive on \(\mathcal{C},\) a semiring, then \(s\) has at most one \(\sigma\)-additive extension to the \(\sigma\)-ring \(\mathcal{R}\) generated by \(\mathcal{C}.\)
(Note that \(s\) is automatically \(\sigma\)-finite if it is finite, e.g., complex or vector valued.)
[Outline: Let
\[s^{\prime}, s^{\prime \prime} : \mathcal{R} \rightarrow E\]
be two \(\sigma\)-additive extensions of \(s.\) By Problem 10, \(\mathcal{R}\) is also generated by
\[\mathcal{C}^{\prime}=\{X \in \mathcal{C}| | s X |<\infty\}.\]
Now set
\[\mathcal{R}^{*}=\left\{X \in \mathcal{R} | s^{\prime} X=s^{\prime \prime} X\right\}.\]
Show that \(\mathcal{R}^{*}\) satisfies properties (i)-(iii) of Theorem 3 in §3, with \(\mathcal{C}\) replaced by \(\mathcal{C}^{\prime};\) so \(\mathcal{R}=\mathcal{R}^{*}\).]
Let \(m_{n}^{*}(n=1,2, \ldots)\) be outer measures in \(S\) such that
\[(\forall X \subseteq S)(\forall n) \quad m_{n}^{*} X \leq m_{n+1}^{*} X.\]
Set
\[\mu^{*}=\lim _{n \rightarrow \infty} m_{n}^{*}.\]
Show that \(\mu^{*}\) is an outer measure in \(S\) (see Note 5).
An outer measure \(m^{*}\) in a metric space \((S, \rho)\) is said to have the Carathéodory property (CP) iff
\[m^{*}(X \cup Y) \geq m^{*} X+m^{*} Y\]
whenever \(\rho(X, Y)>0,\) where
\[\rho(X, Y)=\inf \{\rho(x, y) | x \in X, y \in Y\}.\]
For such \(m^{*},\) prove that
\[m^{*}\left(\bigcup_{k} X_{k}\right)=\sum_{k} m^{*} X_{k}\]
if \(\left\{X_{k}\right\} \subseteq 2^{S}\) and
\[\rho\left(X_{i}, X_{k}\right)>0 \quad(i \neq k).\]
[Hint: For finite unions, use the CP, subadditivity, and induction. Deduce that
\[(\forall n) \sum_{k=1}^{n} m^{*} X_{k} \leq m^{*} \bigcup_{k=1}^{\infty} X_{k}.\]
Let \(n \rightarrow \infty.\) Proceed.]
Let \((S, \mathcal{C}, \mu)\) and \(m^{*}\) be as in Definition 3, with \(\rho\) a metric for \(S.\) Let \(\mu_{n}\) be the restriction of \(\mu\) to the family \(\mathcal{C}_{n}\) of all \(X \in \mathcal{C}\) of diameter
\[d X \leq \frac{1}{n}.\]
Let \(m_{n}^{*}\) be the \(\mu_{n}\)-induced outer measure in \(S.\)
Prove that
(i) \(\left\{m_{n}^{*}\right\} \uparrow\) as in Problem 14;
(ii) the outer measure
\[\mu^{*}=\lim _{n \rightarrow \infty} m_{n}^{*}\]
has the CP (see Problem 15), and
\[\mu^{*} \geq m^{*} \text { on } 2^{S}.\]
[Outline: Let \(\rho(X, Y)>\varepsilon>0(X, Y \subseteq S)\).
If for some \(n, X \cup Y\) has no basic covering from \(\mathcal{C}_{n},\) then
\[\mu^{*}(X \cup Y) \geq m_{n}^{*}(X \cup Y)=\infty \geq \mu^{*} X+\mu^{*} Y,\]
and the CP follows. (Explain!)
Thus assume
\[\left(\forall n>\frac{1}{\varepsilon}\right)(\forall k)\left(\exists B_{n k} \in \mathcal{C}_{n}\right) \quad X \cup Y \subseteq \bigcup_{k=1}^{\infty} B_{n k}.\]
One can choose the \(B_{n k}\) so that
\[\sum_{k=1}^{\infty} \mu B_{n k} \leq m_{n}^{*}(X \cup Y)+\varepsilon.\]
(Why?) As
\[d B_{n k} \leq \frac{1}{n}<\varepsilon,\]
some \(B_{n k}\) cover \(X\) only, others \(Y\) only. (Why?) Deduce that
\[\left(\forall n>\frac{1}{\varepsilon}\right) \quad m_{n}^{*} X+m_{n}^{*} Y \leq \sum_{k=1}^{\infty} \mu_{n} B_{n k} \leq m_{n}^{*}(X \cup Y)+\varepsilon.\]
Let \(\varepsilon \rightarrow 0\) and then \(n \rightarrow \infty\).
Also, \(m^{*} \leq m_{n}^{*} \leq \mu^{*}.\) (Why?)]
Continuing Problem 16, suppose that
\((\forall \varepsilon>0)(\forall n, k)(\forall B \in \mathcal{C})\left(\exists B_{n k} \in \mathcal{C}_{n}\right)\)
\[B \subseteq \bigcup_{k=1}^{\infty} B_{n k} \text { and } \mu B+\varepsilon \geq \sum_{k=1}^{\infty} \mu B_{n k}.\]
Show that
\[m^{*}=\lim _{n \rightarrow \infty} \mu_{n}^{*}=\mu^{*},\]
so \(m^{*}\) itself has the CP.
[Hints: It suffices to prove that \(m^{*} A \geq \mu^{*} A\) if \(m^{*} A<\infty.\) (Why?)
Now, given \(\varepsilon>0, A\) has a covering
\[\left\{B_{i}\right\} \subseteq c\]
such that
\[m^{*} A+\varepsilon \geq \sum \mu B_{i}.\]
(Why?) By assumption,
\[(\forall n) \quad B_{i} \subseteq \bigcup_{k=1}^{\infty} B_{n k}^{i} \in \mathcal{C}_{n} \text { and } \mu B_{i}+\frac{\varepsilon}{2^{i}} \geq \sum_{k=1}^{\infty} \mu B_{n k}^{i}.\]
Deduce that
\[m^{*} A+\varepsilon>\sum \mu B_{i} \geq \sum_{i=1}^{\infty}\left(\sum_{k=1}^{\infty} \mu B_{n k}^{i}-\frac{\varepsilon}{2^{i}}\right)=\sum_{i, k} \mu B_{n k}^{i}-\varepsilon \geq m_{n}^{*} A-\varepsilon.\]
Let \(\varepsilon \rightarrow 0;\) then \(n \rightarrow \infty\).]
Using Problem 17, show that the Lebesgue and Lebesgue-Stieltjes outer measures have the CP.


6.5: Orthogonal Least Squares
\(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \)
\(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \)
Suppose we collect some data when performing an experiment and plot it as shown on the left of Figure 6.5.1. Notice that there is no line on which all the points lie; in fact, it would be surprising if there were since we can expect some uncertainty in the measurements recorded. There does, however, appear to a line, as shown on the right, on which the points almost lie.
In this section, we'll explore how the techniques developed in this chapter enable us to find the line that best approximates the data. More specifically, we'll see how the search for a line passing through the data points leads to an inconsistent system \(A\mathbf x=\mathbf b\text{.}\) Since we are unable to find a solution \(\mathbf x\text{,}\) we instead seek the vector \(\mathbf x\) where \(A\mathbf x\) is as close as possible to \(\mathbf b\text{.}\) Orthogonal projection give us just the right tool for doing this.
Preview Activity 6.5.1.
-
Is there a solution to the equation \(A\mathbf x=\mathbf b\) where \(A\) and \(\mathbf b\) are such that
\begin{equation*} \begin{bmatrix} 1 & 2 \\ 2 & 5 \\ -1 & 0 \\ \end{bmatrix} \mathbf x = \threevec5{-3}{-1}\text{.} \end{equation*}
- We know that \(\threevec12{-1}\) and \(\threevec250\) form a basis for \(Col(A)\text{.}\) Find an orthogonal basis for \(Col(A)\text{.}\)
- Find the orthogonal projection \(\hat{\mathbf{b}}\) of \(\mathbf b\) onto \(Col(A)\text{.}\)
- Explain why the equation \(A\mathbf x=\hat{\mathbf{b}}\) must be consistent and then find its solution.
A first example
When we've encountered inconsistent systems in the past, we've simply said there is no solution and moved on. The preview activity, however, shows how we can find approximate solutions to an inconsistent system: if there are no solutions to \(A\mathbf x = \mathbf b\text{,}\) we instead solve the consistent system \(A\mathbf x = \hat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b\) onto \(Col(A)\text{.}\) As we'll see, this solution is, in a specific sense, the best possible.
Activity 6.5.2.
Suppose we have three data points \((1,1)\text{,}\) \((2,1)\text{,}\) and \((3,3)\) and that we would like to find a line passing through them.
- Plot these three points in Figure 6.5.2. Are you able to draw a line that passes through all three points?
-
Let's write the conditions that would describe a line passing through the points. Remember that the equation of a line can be written as \(b + mx=y\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We will try to find \(b\) and \(m\) so that the three points lie on the line.
The first data point \((1,1)\) gives an equation for \(b\) and \(m\text{.}\) In particular, we know that when \(x=1\text{,}\) then \(y=1\) so we have \(b + m(1) = 1\) or \(b + m = 1\text{.}\) Use the other two data points to create a linear system describing \(m\) and \(b\text{.}\)
-
We have obtained a linear system having three equations, one from each data point, for the two unknowns \(b\) and \(m\text{.}\) Identify a matrix \(A\) and vector \(\mathbf b\) so that the system has the form \(A\mathbf x=\mathbf b\text{,}\) where \(\mathbf x=\twovec bm\text{.}\)
Notice that the unknown vector \(\mathbf x=\twovec bm\) describes the line that we seek.
-
Is there a solution to this linear system? How does this question relate to your attempt to draw a line through the three points above?
- Since this system is inconsistent, we know that \(\mathbf b\) is not in the column space \(Col(A)\text{.}\) Find an orthogonal basis for \(Col(A)\) and use it to find the orthogonal projection \(\hat{\mathbf{b}}\) of \(\mathbf b\) onto \(Col(A)\text{.}\)
- Since \(\hat{\mathbf{b}}\) is in \(Col(A)\text{,}\) the equation \(A\mathbf x = \hat{\mathbf{b}}\) is consistent. Find its solution \(\mathbf x = \twovec{b}{m}\) and sketch the line \(y=b + mx\) in Figure 6.5.2. We say that this is the line of best fit.
This activity illustrates the idea behind a technique known as orthogonal least squares , which we have been working toward throughout this chapter. If the data points are denoted as \((x_i, y_i)\text{,}\) we construct the matrix \(A\) and vector \(\mathbf b\) as
With the vector \(\mathbf x=\twovec bm\) representing the line \(b+mx = y\text{,}\) we see that the equation \(A\mathbf x=\mathbf b\) describes a line passing through all the data points. In our example, it is visually apparent that there is no such line, a fact confirmed by the inconsistency of the equation \(A\mathbf x=\mathbf b\text{.}\)
Remember that \(\hat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b\) onto \(Col(A)\text{,}\) is the closest vector in \(Col(A)\) to \(\mathbf b\text{.}\) Therefore, when we solve the equation \(A\mathbf x=\hat{\mathbf{b}}\text{,}\) we are finding the vector \(\mathbf x\) so that \(A\mathbf x = \threevec{b+mx_1}{b+mx_2}{b+mx_3}\) is as close to \(\mathbf b=\threevec{y_1}{y_2}{y_3}\) as possible. Let's think about what this means within the context of this problem.
The difference \(\mathbf b-A\mathbf x = \threevec{y_1-(b+mx_1)}{y_2-(b+mx_2)}{y_3-(b+mx_3)}\) so that the square of the distance between \(A\mathbf x\) and \(\mathbf b\) is
|\mathbf{b}-A \mathbf{x}^2|=\left(y_1-\left(b+m x_1\right)\right)^2+\left(y_2-\left(b+m x_2\right)\right)^2+\left(y_3-\left(b+m x_3\right)\right)^2 \notag
\]
Our approach finds the values for \(b\) and \(m\) that make this sum of squares as small as possible, which explains why we call this a least squares problem.
Drawing the line defined by the vector \(\mathbf x=\twovec bm\text{,}\) the quantity \(y_i - (b + mx_i)\) reflects the vertical distance between the line and the data point \((x_i, y_i)\text{,}\) as shown in Figure 6.5.5. Seen in this way, the square of the distance \(|{\mathbf b-A\mathbf x}^2| \) is a measure of how much the line defined by the vector \(\mathbf x\) misses the data points. The solution to the least squares problem is the line that misses the data points by the smallest amount possible.
Solving least squares problems
Now that we've seen an example of what we're trying to accomplish, let's put this technique into a more general framework.
Given an inconsistent system \(A\mathbf x = \mathbf b\text{,}\) we seek to find \(\mathbf x\) that minimizes the distance from \(A\mathbf x\) to \(\mathbf b\text{.}\) We find \(\mathbf x\) by forming \(\hat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b\) onto the column space \(Col(A)\) and then solving \(A\mathbf x = \hat{\mathbf{b}}\text{.}\) Moving forward, we will denote the solution of \(A\mathbf x = \hat{\mathbf{b}}\) by \(\hat{\mathbf{x}}\) and call this vector the least squares approximate solution of \(A\mathbf x=\mathbf b\) to distinguish it from a (non-existent) solution of \(A\mathbf x=\mathbf b\text{.}\)
Remember that the orthogonal projection \(\hat{\mathbf{b}}\) of \(\mathbf b\) onto the column space \(Col(A)\) is defined by the fact that \(\hat{\mathbf{b}} - \mathbf b\) is orthogonal to \(Col(A)\text{.}\) In other words, \(\hat{\mathbf{b}}-\mathbf b\) is in the orthogonal complement \(Col(A)^\perp\text{,}\) which Proposition 6.2.10 tells us is the same as \(Nul(A^T)\text{.}\) Since \(\hat{\mathbf{b}}-\mathbf b\) is in \(Nul(A^T)\text{,}\) it follows that
Finally, the least squares approximate solution is the vector \(\hat{\mathbf{x}}\) such that \(A\hat{\mathbf{x}} = \hat{\mathbf{b}}\text{,}\) which gives
Let's record our work in the following proposition.
The least squares approximate solution \(\hat\mathbf x\) to the equation \(A\mathbf x = \mathbf b\) is given by the normal equations
\[ A^TA\hat\mathbf x = A^T\mathbf b\text{.} \]
The linear system represented by the normal equations is consistent since \(\hat{\mathbf{x}}\text{,}\) the least squares approximate solution to \(A\mathbf x=\mathbf b\text{,}\) is a solution. If we further assume that the columns of \(A\) are linearly independent, we can see that there is a unique solution. Imagine, for the moment, that \(\mathbf x\) is a solution to the homogeneous equation \(A^TA\mathbf x = \mathbf 0\text{.}\) We then have
In other words, if \(\mathbf x\) is a solution to the homogeneous equation \(A^TA\mathbf x = \mathbf 0\text{,}\) then we know that \(A\mathbf x = \mathbf 0\text{.}\) Since we are assuming that the columns of \(A\) are linearly independent, we know that the homogeneous equation \(A\mathbf x=\mathbf 0\) has only the zero solution \(\mathbf x = \mathbf 0\text{.}\) Therefore, the homogeneous equation \(A^TA\mathbf x=\mathbf 0\) has only the zero solution, which means that \(A^TA\) has a pivot position in every column. Hence, the normal equations \(A^TA\hat{\mathbf{x}} = A^T\mathbf b\) must have a unique solution.
If the columns of \(A\) are linearly independent, then there is a unique least squares approximate solution \(\hat{\mathbf{x}}\) to the equation \(A\mathbf x=\mathbf b\) given by the normal equations
\[ A^TA\hat{\mathbf{x}} = A^T\mathbf b\text{.} \]
Let's put this proposition to use in the next activity.
Activity 6.5.3.
The rate at which a cricket chirps is related to the outdoor temperature, as reflected in some experimental data that we'll study in this activity. The chirp rate \(C\) is expressed in chirps per second while the temperature \(T\) is in degrees Fahrenheit. Evaluate the following cell to load in the data:
Evaluating this cell also provides:
-
the vectors
chirps
andtemps
formed from the columns of the dataset. -
the command
onesvec(n)
, which creates an \(n\)-dimensional vector whose entries are all one. -
Remember that you can form a matrix whose columns are the vectors
v1
andv2
withmatrix([v1, v2]).T
.
We would like to represent this relationship by a linear function
- Use the first data point \((C_1,T_1)=(20.0,88.6)\) to write an equation involving \(\beta_0\) and \(\beta_1\text{.}\)
-
Suppose that we represent the unknowns using a vector \(\mathbf x = \twovec{\beta_0}{\beta_1}\text{.}\) Use the 15 data points to create the matrix \(A\) and vector \(\mathbf b\) so that the linear system \(A\mathbf x= \mathbf b\) describes the unknown vector \(\mathbf x\text{.}\)
- Write the normal equations \(A^TA\hat{\mathbf{x}} = A^T\mathbf b\text{;}\) that is, find the matrix \(A^TA\) and the vector \(A^T\mathbf b\text{.}\)
-
Solve the normal equations to find \(\hat{\mathbf{x}}\text{,}\) the least squares approximate solution to the equation \(A\mathbf x=\mathbf b\text{.}\) Call your solution
xhat
sincex
has another meaning in Sage.What are the values of \(\beta_0\) and \(\beta_1\) that you found?
-
If the chirp rate is 22 chirps per second, what is your prediction for the temperature?
You can plot the data and your line, assuming you called the solution
xhat
, using the cell below.
This example demonstrates an approach called linear regression , in which a collection of data is modeled using a linear function found by solving a least squares problem. Once we have the linear function that best fits the data, we can make predictions about situations that we haven't encountered in the data.
If we're going to use our function to make predictions, it's natural to ask how much confidence we have in these predictions. This is a statistical question that leads to a rich and well-developed theory, which we won't explore in much detail here. However, there is one simple measure of how well our linear function fits the data that is known as the coefficient of determination and denoted by \(R^2\text{.}\)
We have seen that the square of the distance \(|{\mathbf b-A\mathbf x}^2| \) measures the amount by which the line fails to pass through the data points. When the line is close to the data points, we expect this number to be small. However, the size of this measure depends on the scale of the data. For instance, the two lines shown in Figure 6.5.8 seem to fit the data equally well, but \(|\mathbf b-A\hat{\mathbf{x}}^2| \) is 100 times larger on the right.
The coefficient of determination \(R^2\) is defined by normalizing \(|\mathbf b-A\hat{\mathbf{x}}|^2\) so that it is independent of the scale. Recall that we described how to demean a vector in Section 6.1 : given a vector \(\mathbf v\text{,}\) we obtain \(\widetilde{\mathbf v}\) by subtracting the average of the components from each component.
The coefficient of determination is
where \(\widetilde{\mathbf b}\) is the vector obtained by demeaning \(\mathbf b\text{.}\)
A more complete explanation of this definition relies on the concept of variance, which we explore in Exercise <IP_ADDRESS> and the next chapter. For the time being, it's enough to know that \(0\leq R^2 \leq 1\) and that the closer \(R^2\) is to 1, the better the line fits the data. In our original example, illustrated in Figure 6.5.8, we find that \(R^2 = 0.75\text{,}\) and in our study of cricket chirp rates, we have \(R^2=0.69\text{.}\) However, assessing the confidence we have in predictions made by solving a least squares problem can require considerable thought, and it would be naive to rely only on the value of \(R^2\text{.}\)
Using \(QR\) factorizations
As we've seen, the least squares approximate solution \(\hat{\mathbf{x}}\) to \(A\mathbf x=\mathbf b\) may be found by solving the normal equations \(A^TA\hat{\mathbf{x}} = A^T\mathbf b\text{,}\) and this can be a practical strategy for some problems. However, this approach is not generally sound as small rounding errors can accumulate and lead to inaccurate final results.
As the next activity demonstrates, there is an alternate method for finding the least squares approximate solution \(\hat{\mathbf{x}}\) using a \(QR\) factorization of the matrix \(A\text{,}\) and this method is preferable as it is numerically more reliable.
Activity 6.5.4.
-
Suppose we are interested in finding the least squares approximate solution to the equation \(A\mathbf x = \mathbf b\) and that we have the \(QR\) factorization \(A=QR\text{.}\) Explain why the least squares approximation solution is given by solving
\begin{align*} A\hat{\mathbf{x}} & = QQ^T\mathbf b \\\\ QR\hat{\mathbf{x}} & = QQ^T\mathbf b \\ \end{align*}
-
Multiply both sides of the second expression by \(Q^T\) and explain why
\begin{equation*} R\hat{\mathbf{x}} = Q^T\mathbf b. \end{equation*}
Since \(R\) is upper triangular, this is a relatively simple equation to solve using back substitution, as we saw in Section 5.1 . We will therefore write the least squares approximate solution as
\begin{equation*} \hat{\mathbf{x}} = R^{-1}Q^T\mathbf b, \end{equation*}and put this to use in the following context.
-
Brozak’s formula, which is used to calculate a person's body fat index \(BFI\text{,}\) is
\begin{equation*} BFI = 100 \left(\frac{4.57}{\rho} - 4.142\right) \end{equation*}
where \(\rho\) denotes a person's body density in grams per cubic centimeter. Obtaining an accurate measure of \(\rho\) is difficult, however, because it requires submerging the person in water and measuring the volume of water displaced. Instead, we will gather several other body measurements, which are more easily obtained, and use it to predict \(BFI\text{.}\)
For instance, suppose we take 10 patients and measure their weight \(w\) in pounds, height \(h\) in inches, abdomen \(a\) in centimeters, wrist circumference \(r\) in centimeters, neck circumference \(n\) in centimeters, and \(BFI\text{.}\) Evaluating the following cell loads and displays the data.
In addition, that cell provides:-
vectors
weight
,height
,abdomen
,wrist
,neck
, andBFI
formed from the columns of the dataset. -
the command
onesvec(n)
, which returns an \(n\)-dimensional vector whose entries are all one. -
the command
QR(A)
that returns the \(QR\) factorization of \(A\) asQ, R = QR(A)
. -
the command
demean(v)
, which returns the demeaned vector \(\widetilde{\mathbf v}\text{.}\)
We would like to find the linear function
\begin{equation*} \beta_0 + \beta_1w + \beta_2h + \beta_3a + \beta_4r + \beta_5n = BFI \end{equation*}that best fits the data.
Use the first data point to write an equation for the parameters \(\beta_0,\beta_1,\ldots,\beta_5\text{.}\)
-
vectors
- Describe the linear system \(A\mathbf x = \mathbf b\) for these parameters. More specifically, describe how the matrix \(A\) and the vector \(\mathbf b\) are formed.
-
Construct the matrix \(A\) and find its \(QR\) factorization in the cell below.
-
Find the least squares approximate solution \(\hat{\mathbf{x}}\) by solving the equation \(R\hat{\mathbf{x}} = Q^T\mathbf b\text{.}\) You may want to use
N(xhat)
to display a decimal approximation of the vector. What are the parameters \(\beta_0,\beta_1,\ldots,\beta_5\) that best fit the data? -
Find the coefficient of determination \(R^2\) for your parameters. What does this imply about the quality of the fit?
- Suppose a person's measurements are: weight 190, height 70, abdomen 90, wrist 18, and neck 35. Estimate this person's \(BFI\text{.}\)
To summarize, we have seen that
If the columns of \(A\) are linearly independent and we have the \(QR\) factorization \(A=QR\text{,}\) then the least squares approximate solution \(\hat{\mathbf{x}}\) to the equation \(A\mathbf x=\mathbf b\) is given by
Polynomial Regression
In the examples we've seen so far, we have fit a linear function to a dataset. Sometimes, however, a polynomial, such as a quadratic function, may be more appropriate. It turns out that the techniques we've developed in this section are still useful as the next activity demonstrates.
Activity 6.5.5.
-
Suppose that we have a small dataset containing the points \((0,2)\text{,}\) \((1,1)\text{,}\) \((2,3)\text{,}\) and \((3,3)\text{,}\) such as appear when the following cell is evaluated.
In addition to loading and plotting the data, evaluating that cell provides the following commands:
-
Q, R = QR(A)
returns the \(QR\) factorization of \(A\text{.}\) -
demean(v)
returns the demeaned vector \(\widetilde{\mathbf v}\text{.}\)
Let's fit a quadratic function of the form
\begin{equation*} \beta_0 + \beta_1 x + \beta_2 x^2 = y \end{equation*}to this dataset.
Write four equations, one for each data point, that describe the coefficients \(\beta_0\text{,}\) \(\beta_1\text{,}\) and \(\beta_2\text{.}\)
-
-
Express these four equations as a linear system \(A\mathbf x = \mathbf b\) where \(\mathbf x = \threevec{\beta_0}{\beta_1}{\beta_2}\text{.}\)
Find the \(QR\) factorization of \(A\) and use it to find the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\)
-
Use the parameters \(\beta_0\text{,}\) \(\beta_1\text{,}\) and \(\beta_2\) that you found to write the quadratic function that fits the data. You can plot this function, along with the data, by entering your function in the appropriate place below.
- What is your predicted \(y\) value when \(x=1.5\text{.}\)
- Find the coefficient of determination \(R^2\) for the quadratic function? What does this say about the quality of the fit?
-
Now fit a cubic polynomial of the form
\begin{equation*} \beta_0 + \beta_1x + \beta_2 x^2 + \beta_3x^3 = y \end{equation*}
to this dataset.
- Find the coefficient of determination \(R^2\) for the cubic function. What does this say about the quality of the fit?
-
What do you notice when you plot the cubic function along with the data? How does this reflect the value of \(R^2\) that you found?
The matrices \(A\) that you created in the last activity when fitting a quadratic and cubic function to a dataset have a special form. In particular, if the data points are labeled \((x_i, y_i)\) and we seek a degree \(k\) polynomial, then
This is called a Vandermonde matrix of degree \(k\text{.}\)
Activity 6.5.6.
This activity explores a dataset describing Arctic sea ice and that comes from Sustainability Math.
Evaluating the cell below will plot the extent of Arctic sea ice, in millions of square kilometers, during the twelve months of 2012.
In addition, you have access to a few special variables and commands:
-
month
is the vector of month values andice
is the vector of sea ice values from the table above. -
vandermonde(x, k)
constructs the Vandermonde matrix of degree \(k\) using the points in the vectorx
. -
Q, R = QR(A)
provides the \(QR\) factorization of \(A\text{.}\) -
demean(v)
returns the demeaned vector \(\widetilde{\mathbf v}\text{.}\)
-
Find the vector \(\hat{\mathbf{x}}\text{,}\) the least squares approximate solution to the linear system that results from fitting a degree 5 polynomial to the data.
-
If your result is stored in the variable
xhat
, you may plot the polynomial and the data together using the following cell. - Find the coefficient of determination \(R^2\) for this polynomial fit.
-
Repeat these steps to fit a degree 8 polynomial to the data, plot the polynomial with the data, and find \(R^2\text{.}\)
-
Repeat one more time by fitting a degree 11 polynomial to the data, plotting it, and finding \(R^2\text{.}\)
It's certainly true that higher degree polynomials fit the data better, as seen by the increasing values of \(R^2\text{,}\) but that's not always a good thing. For instance, when \(k=11\text{,}\) you may notice that the graph of the polynomial wiggles a little more than we would expect. In this case, the polynomial is trying too hard to fit the data, which usually contains some uncertainty, especially if it's obtained from measurements. The error built in to the data is called noise, and its presence means that we shouldn't expect our polynomial to fit the data perfectly. When we choose a polynomial whose degree is too high, we give the noise too much weight in the model, which leads to some undesirable behavior, like the wiggles in the graph.
Fitting the data with a polynomial whose degree is too high is called overfitting , a phenomenon that can appear in many machine learning applications. Generally speaking, we would like to choose \(k\) large enough to capture the essential features of the data but not so large that we overfit and build the noise into the model. There are ways to determine the optimal value of \(k\text{,}\) but we won't pursue that here.
- Choosing a reasonable value of \(k\text{,}\) estimate the extent of Arctic sea ice at month 6.5, roughly at the Summer Solstice.
Summary
This section introduced some types of least squares problems and a framework for working with them.
- Given an inconsistent system \(A\mathbf x=\mathbf b\text{,}\) we find \(\hat{\mathbf{x}}\text{,}\) the least squares approximate solution, by requiring that \(A\hat{\mathbf{x}}\) be as possible to \(\mathbf b\) as possible. In other words, \(A\hat{\mathbf{x}} = \hat{\mathbf{b}}\) where \(\hat{\mathbf{b}}\) is the orthogonal projection of \(\mathbf b\) onto \(Col(A)\text{.}\)
- One way to find \(\hat{\mathbf{x}}\) is by solving the normal equations \(A^TA\hat{\mathbf{x}} = A^T\mathbf b.\) This is not our preferred method since numerical problems can arise.
- A second way to find \(\hat{\mathbf{x}}\) uses a \(QR\) factorization of \(A\text{.}\) If \(A=QR\text{,}\) then \(\hat{\mathbf{x}} = R^{-1}Q^T\mathbf b\) and finding \(R^{-1}\) is computationally feasible since \(R\) is upper triangular.
- This technique may be applied widely and is useful for modeling data. We saw examples in this section where linear functions of several input variables and polynomials provided effective models for different datasets.
- A simple measure of the quality of the fit is the coefficient of determination \(R^2\) though some additional thought should be given in real applications.
Exercises 6.5.6Exercises
Evaluating the following cell loads in some commands that will be helpful in the following exercises. In particular, there are commands
-
QR(A)
that returns the \(QR\) factorization ofA
asQ, R = QR(A)
, -
onesvec(n)
that returns the \(n\)-dimensional vector whose entries are all 1, -
demean(v)
that demeans the vectorv
, -
vandermonde(x, k)
that returns the Vandermonde matrix of degree \(k\) formed from the components of the vectorx
, and -
plot_model(xhat, data)
that plots thedata
and the modelxhat
.
Suppose we write the linear system
as \(A\mathbf x=\mathbf b\text{.}\)
- Find an orthogonal basis for \(Col(A)\text{.}\)
- Find \(\hat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b\) onto \(Col(A)\text{.}\)
- Find a solution to the linear system \(A\mathbf x = \hat{\mathbf{b}}\text{.}\)
Consider the data in Table 6.5.11.
| \(x\) | \(y\) |
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
- Set up the linear system \(A\mathbf x=\mathbf b\) that describes the line \(b + mx = y\) passing through these points.
- Write the normal equations that describe the least squares approximate solution to \(A\mathbf x=\mathbf b\text{.}\)
- Find the least squares approximate solution \(\hat{\mathbf{x}}\) and plot the data and the resulting line.
- What is your predicted \(y\)-value when \(x=3.5\text{?}\)
- Find the coefficient of determination \(R^2\text{.}\)
Consider the four points in Table 6.5.11.
-
Set up a linear system \(A\mathbf x = \mathbf b\) that describes a quadratic function
\begin{equation*} \beta_0+\beta_1x+\beta_2x^2 = y \end{equation*}
passing through the points.
- Use a \(QR\) factorization to find the least squares approximate solution \(\hat{\mathbf{x}}\) and plot the data and the graph of the resulting quadratic function.
- What is your predicted \(y\)-value when \(x=3.5\text{?}\)
- Find the coefficient of determination \(R^2\text{.}\)
Consider the data in Table 6.5.12.
| \(x_1\) | \(x_2\) | \(y\) |
| 1 | 1 | 4.2 |
| 1 | 2 | 3.3 |
| 2 | 1 | 5.9 |
| 2 | 2 | 5.1 |
| 3 | 2 | 7.5 |
| 3 | 3 | 6.3 |
-
Set up a linear system \(A\mathbf x = \mathbf b\) that describes the relationship
\begin{equation*} \beta_0 + \beta_1 x_1 + \beta_2 x_2 = y. \end{equation*}
- Find the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\)
- What is your predicted \(y\)-value when \(x_1 = 2.4\) and \(x_2=2.9\text{?}\)
- Find the coefficient of determination \(R^2\text{.}\)
Determine whether the following statements are true or false and explain your thinking.
- If \(A\mathbf x=\mathbf b\) is consistent, then \(\hat{\mathbf{x}}\) is a solution to \(A\mathbf x=\mathbf b\text{.}\)
- If \(R^2=1\text{,}\) then the least squares approximate solution \(\hat{\mathbf{x}}\) is also a solution to the original equation \(A\mathbf x=\mathbf b\text{.}\)
- Given the \(QR\) factorization \(A=QR\text{,}\) we have \(A\hat{\mathbf{x}}=Q^TQ\mathbf b\text{.}\)
- A \(QR\) factorization provides a method method for finding the approximate least squares solution to \(A\mathbf x=\mathbf b\) that is more reliable than solving the normal equations.
- A solution to \(AA^T\mathbf x = A\mathbf b\) is the least squares approximate solution to \(A\mathbf x = \mathbf b\text{.}\)
Explain your response to the following questions.
- If \(\hat{\mathbf{x}}=\mathbf 0\text{,}\) what does this say about the vector \(\mathbf b\text{?}\)
- If the columns of \(A\) are orthonormal, how can you easily find the least squares approximate solution to \(A\mathbf x=\mathbf b\text{?}\)
The following cell loads in some data showing the number of people in Bangladesh living without electricity over 27 years. It also defines vectors
year
, which records the years in the data set, and
people
, which records the number of people.
-
Suppose we want to write
\begin{equation*} N = \beta_0 + \beta_1 t \end{equation*}
where \(t\) is the year and \(N\) is the number of people. Construct the matrix \(A\) and vector \(\mathbf b\) so that the linear system \(A\mathbf x=\mathbf b\) describes the vector \(\mathbf x=\twovec{\beta_0}{\beta_1}\text{.}\)
- Using a \(QR\) factorization of \(A\text{,}\) find the values of \(\beta_0\) and \(\beta_1\) in the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\)
- What is the coefficient of determination \(R^2\) and what does this tell us about the quality of the approximation?
- What is your prediction for the number of people living without electricity in 1985?
- Estimate the year in which there will be no people living without electricity.
This problem concerns a data set describing planets in our Solar system. For each planet, we have the length \(L\) of the semi-major axis, essentially the distance from the planet to the Sun in AU (astronomical units), and the period \(P\text{,}\) the length of time in years required to completed one orbit around the Sun.
We would like to model this data using the function \(P = CL^r\) where \(C\) and \(r\) are parameters we need to determine. Since this isn't a linear function, we will transform this relationship by taking the natural logarithm of both sides to obtain
Evaluating the following cell loads the data set and defines two vectors
logaxis
, whose components are \(\ln(L)\text{,}\) and
logperiod
, whose components are \(\ln(P)\text{.}\)
- Construct the matrix \(A\) and vector \(\mathbf b\) so that the solution to \(A\mathbf x=\mathbf b\) is the vector \(\mathbf x=\twovec{\ln(C)}r\text{.}\)
- Find the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\) What does this give for the values of \(C\) and \(r\text{?}\)
- Find the coefficient of determination \(R^2\text{.}\) What does this tell us about the quality of the approximation?
- Suppose that the orbit of an asteroid has a semi-major axis whose length is \(L=4.0\) AU. Estimate the period \(P\) of the asteroid's orbit.
- Halley's Comet has a period of \(P=75\) years. Estimate the length of its semi-major axis.
Evaluating the following cell loads a data set describing the temperature in the Earth's atmosphere at various altitudes. There are also two vectors
altitude
, expressed in kilometers, and
temperature
, in degress Celsius.
- Describe how to form the matrix \(A\) and vector \(\mathbf b\) so that the linear system \(A\mathbf x=\mathbf b\) describes a degree \(k\) polynomial fitting the data.
- Choose a value of \(k\text{,}\) construct the matrix \(A\) and vector \(\mathbf b\text{,}\) and find the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\)
-
Plot the polynomial and data using
plot_model(xhat, data)
. - Now examine what happens as you vary the degree of the polynomial \(k\text{.}\) Choose an appropriate value of \(k\) that seems to capture the most important features of the data while avoiding overfitting, and explain your choice.
- Use your value of \(k\) to estimate the temperature at an altitude of 55 kilometers.
The following cell loads some data describing 1057 houses in a particular real estate market. For each house, we record the living area in square feet, the lot size in acres, the age in years, and the price in dollars. The cell also defines variables
area
,
size
,
age
, and
price
.
- Use a \(QR\) factorization to find the least squares approximate solution \(\hat{\mathbf{x}}\text{.}\)
- Discuss the significance of the signs of \(\beta_1\text{,}\) \(\beta_2\text{,}\) and \(\beta_3\text{.}\)
- If two houses are identical except for differing in age by one year, how would you predict their prices compare to one another?
- Find the coefficient of determination \(R^2\text{.}\) What does this say about the quality of the fit?
- Predict the price of a house whose living area is 2000 square feet, lot size is 1.5 acres, and age is 50 years.
This problem is about the meaning of the coefficient of determination \(R^2\) and its connection to variance, a topic that appears in the next section. Throughout this problem, we consider the linear system \(A \mathbf{x}=\mathbf{b}\) and the approximate least-squares solution \(\widehat{\mathbf{x}}\), where \(A \widehat{\mathbf{x}}=\widehat{\mathbf{b}}\). We suppose that \(A\) is an \(m \times n\) matrix, and we will denote the \(m\)-dimensional vector \(\mathbf{1}=\left[\begin{array}{c}1 \\ 1 \\ \vdots \\ 1\end{array}\right]\).
-
Explain why \(\overline{\mathbf{b}}\), the mean of the components of \(\mathbf{b}\), can be found as the dot product \[\overline{\mathbf{b}}=\frac{1}{m} \mathbf{b} \cdot \mathbf{1}\]
-
In the examples we have seen in this section, explain why \(\mathbf{1} \) is in \(Col(A)\text{.}\)
-
If we write \(\mathbf b = \hat{\mathbf{b}} + \mathbf b^\perp\text{,}\) we explain why
\begin{equation*} \mathbf b^\perp\cdot\mathbf{1} = 0 \end{equation*}
and hence why the mean of the components of \(\mathbf b^\perp\) is zero.
-
The variance of an \(m\)-dimensional vector \(\mathbf v\) is \(Var(\mathbf v) = \frac1m |{\widetilde{\mathbf v}}^2 |\text{,}\) where \(\widetilde{\mathbf v}\) is the vector obtained by demeaning \(\mathbf v\text{.}\)
-
Explain why
\begin{equation*} Var(\mathbf b) = Var(\hat{\mathbf{b}}) + Var(\mathbf b^\perp). \end{equation*}
-
Explain why
\begin{equation*} \frac{| {\mathbf b - A\hat{\mathbf{x}}}^2}| {|{\widetilde{\mathbf b}}^2}| = \frac{Var(\mathbf b^\perp)}{Var(\mathbf b)} \end{equation*}
and hence
\begin{equation*} R^2 = \frac{Var(\hat{\mathbf{b}})}{Var(\mathbf b)} = \frac{Var(A\hat{\mathbf{x}})}{Var(\mathbf b)}. \end{equation*}
These expressions indicate why it is sometimes said that \(R^2\) measures the “fraction of variance explained” by the function we are using to fit the data. As seen in the previous exercise, there may be other features that are not recorded in the dataset that influence the quantity we wish to predict.
- Explain why \(0\leq R^2 \leq 1\text{.}\)